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Geometry Level 3

In a Δ A B C \Delta ABC , the altitude, the angle bisector and the median from A A divides the ∠ A \angle A into four equal angles. Find the value of ∣ ∠ B − ∠ C ∣ \bigg| \angle B-\angle C \bigg| in degrees.

3 0 ∘ 30^\circ 67. 5 ∘ 67.5^\circ 4 5 ∘ 45^\circ 6 0 ∘ 60^\circ 22. 5 ∘ 22.5^\circ 37. 5 ∘ 37.5^\circ 7 5 ∘ 75^\circ 1 5 ∘ 15^\circ

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Satyajit Mohanty by Satyajit Mohanty , 24, India

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1 solution

L e t ∠ A = 4 α . ∴ ∠ B = 90 − α . ∠ C = 90 − 3 α . I n Δ A B F , A F S i n B = B F S i n 3 α ⟹ A F C o s α = B C 2 ∗ S i n 3 α . . . . . ( ∗ ) I n Δ A F C , A F S i n C = F C S i n α ⟹ A F C o s 3 α = B C 2 ∗ S i n α . . . . . ( ∗ ∗ ) ∴ f r o m ( ∗ ) a n d ( ∗ ∗ ) S i n 2 α = S i n 6 α . B u t S i n 45 = S i n ( 3 ∗ 45 ) ∴ ∣ ∠ B − ∠ C ∣ = 2 α = 45. Let~ \angle A= 4\alpha.~~~~~\therefore~\angle B=90-\alpha.~~~~~\angle C=90 - 3\alpha.\\ In~\Delta ABF, ~~\dfrac{AF}{SinB}=\dfrac{BF}{Sin3\alpha} ~~~~~\implies~\dfrac{AF}{Cos\alpha}=\dfrac{BC}{2*Sin3\alpha}.....(*)\\ In~\Delta AFC, ~~\dfrac{AF}{SinC}=\dfrac{FC}{Sin\alpha} ~~~~~\implies~\dfrac{AF}{Cos3\alpha}=\dfrac{BC}{2*Sin\alpha}.....(**)\\ \therefore~ from~(*)~and~(**) ~~Sin2\alpha=Sin6\alpha .\\ But~Sin45=Sin(3*45)~~~~\therefore~|\angle B- \angle C|=2\alpha=~~~~\Huge \color{#D61F06}{45}.

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